Is ${298473}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {298473}= &&{2}\cdot100000+ \\&&{9}\cdot10000+ \\&&{8}\cdot1000+ \\&&{4}\cdot100+ \\&&{7}\cdot10+ \\&&{3}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {298473}= &&{2}(99999+1)+ \\&&{9}(9999+1)+ \\&&{8}(999+1)+ \\&&{4}(99+1)+ \\&&{7}(9+1)+ \\&&{3} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {298473}= &&\gray{2\cdot99999}+ \\&&\gray{9\cdot9999}+ \\&&\gray{8\cdot999}+ \\&&\gray{4\cdot99}+ \\&&\gray{7\cdot9}+ \\&& {2}+{9}+{8}+{4}+{7}+{3} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${298473}$ is divisible by $3$ if ${ 2}+{9}+{8}+{4}+{7}+{3}$ is divisible by $3$ Add the digits of ${298473}$ $ {2}+{9}+{8}+{4}+{7}+{3} = {33} $ If ${33}$ is divisible by $3$ , then ${298473}$ must also be divisible by $3$ ${33}$ is divisible by $3$, therefore ${298473}$ must also be divisible by $3$.